poj 1789 Truck History

 where the sum goes over all pairs of types in the derivation plan such that t_o is the original type and t_d the type derived from it and d(t_o,t_d) is the distance of the types

1/Σ_(to,td)d(t_o,t_d)

派生關係有向邊，邊權爲兩個串的不同字符的個數，要求的是找出一種用到所有串的派生關係，使得Q最小，（因爲是1/Q），所以這題就是求最小生成樹

 

最樸素的最小生成樹AC

 

#include<stdio.h> const int INF = 0x3f3f3f3f; const int MAXN = 2000; char word[MAXN][8]; int map[MAXN][MAXN]; int n, lowcost[MAXN], cloest[MAXN]; bool used[MAXN]; inline int countDis(int i, int j){ int cnt = 0; for(int k = 0; k < 7; ++k){ if(word[i][k] != word[j][k]) cnt++; } return cnt; } int prim(){ int ans = 0, min, minidx; for(int i = 0; i < n; ++i){ lowcost[i] = map[0][i]; cloest[i] = 0; used[i] = false; } used[0] = true; for(int i = 1; i < n; ++i){ min = INF; for(int j = 0; j < n; ++j){ if(!used[j] && lowcost[j] < min) min = lowcost[j], minidx = j; } used[minidx] = true; ans += min; for(int j = 0; j < n; ++j){ if(!used[j] && lowcost[j] > map[minidx][j]) lowcost[j] = map[minidx][j], cloest[j] = minidx; } } return ans; } int main(){ while(scanf("%d",&n), n){ for(int i = 0; i < n; ++i){ scanf("%s",word[i]); } for(int i = 0; i < n; ++i){ for(int j = 0; j < n; ++j){ map[i][j] = countDis(i, j); } } printf("The highest possible quality is 1/%d./n",prim()); } return 0; }